How do I use Python to pack a big integer treating four characters as an unsigned long in network byte order as Ruby's .pack("N") does?

This is the bit of Ruby I want to implement in Python:

Base64.urlsafe_encode64([Digest::MD5.hexdigest(url).to_i(16)].pack("N")).sub(/==\n?$/, '')

You see, this helps turn a URL like this:

http://stackoverflow.com/questions/ask

Into a small code like thise:

sUEBtw

The big integer that gets generated in the process is this:

307275247029202263937236026733300351415

I've been able to pack this into binary form using this Python code:

url = 'http://stackoverflow.com/questions/ask'
n = int(hashlib.md5(url).hexdigest(), 16)                                       
s = struct.Struct('d')                                                          
values = [n]                                                                    
packed_data = s.pack(*values)                                                   
short_code = base64.urlsafe_b64encode(packed_data)[:-1]
print short_code

The short code I get is this:

zgMM62Hl7Ec

As you can see it's larger than the one I get with Ruby this the packing is using a different format.

You're help will be appreciated.

Solution

This does the trick:

import hashlib
import base64
url = 'http://stackoverflow.com/questions/ask'
print base64.urlsafe_b64encode(hashlib.md5(url).digest()[-4:])[:-2]

Output

sUEBtw

.digest() gives the packed bytes of the full 16-byte digest so no need for struct.pack, but it seems Ruby's .pack('N') only converts the last four bytes of the digest.